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    (Original post by lefterispower)
    I'm a little confused about finding dx/dt of x = sin^3t...any help pls? )
    Think chain rule... let u=sint
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    (Original post by lefterispower)
    I'm a little confused about finding dx/dt of x = sin^3t...any help pls? )
     x = sin^3 t

    is the same as  x = (sint)^3

    then you would do the chain rule to get dx/dt
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    (Original post by Zaphod77)
    Think chain rule... let u=sint
    thanks...!!
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    (Original post by masryboy94)
     x = sin^3 t

    is the same as  x = (sint)^3

    then you would do the chain rule to get dx/dt
    thanks bro!!
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    (Original post by masryboy94)
    shouldn't the second one's denominator be x + 3? and i don't know how to do the last 2 ... THANK YOU SO MUCH BTW !
    Yes it should. My mistake. Fixed it now though.

    For \frac{x+2}{x-2}, you want to write x+2 into something that has x-2. You can't do anything with the x, but you can rewrite the +2 differently. Think of two numbers that make +2. From the bottom of the fraction you know one number is -2. What is the other?

    edit: Perhaps a better way to think of it is like this: The aim to to eventually cancel out the x-2 on the bottom with x-2 on the top to get 1. So you know the top has to be written as: (x-2 + \mathrm{something}), where the something has to be set so that the top is still x+2, but written differently.

    The last one follows similarly.

    Give this a go too:
    \dfrac{2x-1}{x-1}

    Spoiler:
    Show
    \dfrac{2x-1}{x-1} = \dfrac{x+x-1}{x-1} = \dfrac{x}{x-1}+1 = \dfrac{x-1+1}{x-1} + 1 = 1+ \dfrac{1}{x-1}+1 = 2+ \dfrac{1}{x-1}
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    (Original post by F1Addict)
    Yes it should. My mistake. Fixed it now though.

    For \frac{x+2}{x-2}, you want to write x+2 into something that has x-2. You can't do anything with the x, but you can rewrite the +2 differently. Think of two numbers that make +2. From the bottom of the fraction you know one number is -2. What is the other?

    The last one follows similarly.
     \frac{x+4-2}{x-2} ?
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    (Original post by masryboy94)
     x = sin^3 t

    is the same as  x = (sint)^3

    then you would do the chain rule to get dx/dt
    We were told to split it to sin(x) x sin^2(x)
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    (Original post by physicso)
    Can someone please explain to me why y=1-2cosx intersects at pi/3 and 5pi/3? surely it should be pi/3 and (pi/3 + pi)= 4pi/3?
    draw a graph, here, this might help you:

    sorry for the scrunched paper,

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    Although I would presume that it is a very rare occurrence, has anybody heard of anyone getting full UMS on C4?
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    (Original post by masryboy94)
     \frac{x+4-2}{x-2} ?
    Thats correct, now how can you simplify it further? (Note that x+4-2 is the same as x-2+4).

    Check my previous post as I edited it to give another, slightly different, way to think of it. Also added a harder fraction problem which you should try.
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    (Original post by BearOfStown)
    We were told to split it to sin(x) x sin^2(x)
    you would if you were integrating, but for differentiating you can do the chain rule
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    (Original post by überambitious_ox)
    Although I would presume that it is a very rare occurrence, has anybody heard of anyone getting full UMS on C4?
    Not yet but there's always a first time and thats what we should be aiming for
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    How likely is it that there's an integration question which can only be solved by substitution and they don't tell you that substitution values? This is something I'm genuinely worried about :/
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    (Original post by überambitious_ox)
    Although I would presume that it is a very rare occurrence, has anybody heard of anyone getting full UMS on C4?
    Yep, someone in my school last year managed it
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    guys for edexcel c4, what are the most difficult of the standard edexcel or solomon c4 papers from experience or grade boundaries
    thanks
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    (Original post by Chris-69)
    Can only help me understand what I'm supposed to be doing in this question:

    "At a given instant, the radii of two concentric circles are 8cm and 12cm. The radius of the outer circle is increasing at a rate of 1cm/s and the radius of the inner circle is increasing at a rate of 2cm/s.

    Find the rate of change of the area enclosed by the two circles."

    Thanks!
    Help anyone?
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    (Original post by F1Addict)
    Yes it should. My mistake. Fixed it now though.

    For \frac{x+2}{x-2}, you want to write x+2 into something that has x-2. You can't do anything with the x, but you can rewrite the +2 differently. Think of two numbers that make +2. From the bottom of the fraction you know one number is -2. What is the other?

    edit: Perhaps a better way to think of it is like this: The aim to to eventually cancel out the x-2 on the bottom with x-2 on the top to get 1. So you know the top has to be written as: (x-2 + \mathrm{something}), where the something has to be set so that the top is still x+2, but written differently.

    The last one follows similarly.

    Give this a go too:
    \dfrac{2x-1}{x-1}

    Spoiler:
    Show
    \dfrac{2x-1}{x-1} = \dfrac{x+x-1}{x-1} = \dfrac{x}{x-1}+1 = \dfrac{x-1+1}{x-1} + 1 = 1+ \dfrac{1}{x-1}+1 = 2+ \dfrac{1}{x-1}
    Using long division you get 2 + 2/x -1?????????????

    whoopsie i used 2x +1 instead of 2x-1
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    http://www.edexcel.com/migrationdocu...e_20110620.pdf

    For question 6b of that paper... I got the angle of 110.9 (which is not acute). Why do you have to do 180 - 110.9 to find the acute angle? Thanks
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    (Original post by bestfriends33)
    http://www.edexcel.com/migrationdocu...e_20110620.pdf

    For question 6b of that paper... I got the angle of 110.9 (which is not acute). Why do you have to do 180 - 110.9 to find the acute angle? Thanks
    draw the lines intersecting and you can see which angle you have just worked out of 111 between the lines, then you can see there is another angle which is also between the lines but acute
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    (Original post by bestfriends33)
    http://www.edexcel.com/migrationdocu...e_20110620.pdf

    For question 6b of that paper... I got the angle of 110.9 (which is not acute). Why do you have to do 180 - 110.9 to find the acute angle? Thanks
    What I usually do is, if I get a negative value when doing the scalar product, just take the positive value of it and you'll get the acute angle.
 
 
 
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